In a Young’s double slit experiment, light of wavelength 600 nm is used. Slits are 1 mm apart and fringes are observed on a screen 2.0 m away from the slits. The minimum distance from the centre of the screen where intensity of fringe is half of the maximum intensity is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.3 mm
In Young’s experiment the intensity at a point depends on the phase difference \( \phi \) between the two waves.
\( \displaystyle I = I_0\cos^{2}\frac{\phi}{2} \)
For half the maximum intensity:
\( \displaystyle \cos^{2}\frac{\phi}{2} = \frac{1}{2} \)
\( \displaystyle \frac{\phi}{2} = \frac{\pi}{4} \;\Rightarrow\; \phi = \frac{\pi}{2} \)
The corresponding path difference is
\( \displaystyle \Delta x = \frac{\lambda}{2\pi}\,\phi = \frac{\lambda}{4} \)
At a distance y from the centre the path difference is \( yd/D \).
\( \displaystyle \frac{yd}{D} = \frac{\lambda}{4} \;\Rightarrow\; y = \frac{\lambda D}{4d} \)
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( d = 1\,\mathrm{mm} = 1\times10^{-3}\,\mathrm{m} \)
\( \displaystyle y = \frac{6\times10^{-7}\,\mathrm{m} \times 2.0\,\mathrm{m}}{4 \times 1\times10^{-3}\,\mathrm{m}} \)
\( y = 3\times10^{-4}\,\mathrm{m} = 0.3\,\mathrm{mm} \)
Shortcut: The fringe width is \( \beta = \lambda D/d = 1.2 \) mm, and half the maximum intensity occurs a quarter of a fringe width from the centre: \( \beta/4 = 0.3 \) mm.